UESTC 1652

題意：中文題

思路：遍歷每一公裏，然後計算每個車道對後一公裏做出的貢獻，最邊上的車道特判，遍歷完所有車道後判斷如果車道上有障礙，那麽這公裏的的這個車道的概率為0，數據比較水，如果數據大可以用矩陣快速冪對每2個障礙之間用矩陣快速冪優化，這裏就不寫了

AC代碼：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include 
 
"algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
 
const int N=1e5+100;
const ll mod=1e9+7;

double dp[2][30005],ans;
int ma,n,m,k,a,b,p;
vector<int> pp[1005];
int main(){
    //ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>m>>k>>n>>p;
    for(int i=1; i<=k; ++i){
        cin>>a>>b;
        pp[b].pb(a);
        ma
 
=max(ma,b);
    }
    dp[0][p]=1;
    double *now=dp[0], *nex=dp[1];
    for(int i=1; i<=ma; ++i){
        for(int j=1; j<=m; ++j){
            if(j==1){
                nex[j]+=now[j]/2;
                nex[j+1]+=now[j]/2;
            }
            else if(j==m){
                nex[j]+=now[j]/2;
                nex[j-1]+=now[j]/2;
            }
            else{
                nex[j]+=now[j]/3;
                nex[j+1]+=now[j]/3;
                nex[j-1]+=now[j]/3;
            }
        }
        for(auto j : pp[i]){
            nex[j]=0;
        }
        swap(now,nex);
        memset(nex,0,sizeof(dp[0]));
    }
    for(int i=1; i<=m; ++i){
        ans+=now[i];
    }
    printf("%.6f\n",ans);
    return 0;
}

UESTC 電子科大專題訓練 DP-E